Q73 / 155 · WBJEE 2024 · Mathematics

If $\left(1+x+x^2+x^3\right)^5=\sum_{k=0}^{15} a_k x^k$ then $\sum_{k=0}^7(-1)^k \cdot a_{2 k}$ is equal to

  • A
    $2^5$
  • B
    $4^5$
  • Correct answer
    C
    0
  • D
    $4^4$

Solution

$\begin{aligned} & \text { Hint : } \sum_{k=0}^1(-1)^k a_{2 k} \\ & =a_0-a_2+a_4-a_6+a_8 \ldots . a_{14} \\ & \left(1+x+x^2+x^3\right)^5=a_0+a_1 x+\ldots . .+a_{15} x^{15} \\ & \text { put } x=i \\ & \left(1+i+i^2+i^3\right)^5=a_0+a_1 i+a_2 i^2+\ldots .+a_{15} i^{15} \end{aligned}$ equating real part $a_0-a_2+a_4-a_6+a_8 \ldots \ldots . a_{14}=0$

Hence, the correct answer is option: 'C'.

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